[weinzierl]

Excellent answer and I am sure you are aware of this, but like to point out:

"Mathematically, the Fourier transform is "simply" a way of representing time signals in a certain orthogonal vectorial basis."

Not just time signals but any piecewise continuous and differentiable as well as Dirichlet integrable function. This has many applications, just a few examples from the top of my head: image processing, solving differential equations, fast multiplication.

I'd also like to add that from a mathematical point of view these transforms are "lossless" in the sense that the transformed function has the exact same information as the original and you can get back the exact original even if all you have is the transform.

I feel this often gets lost when people approach the Fourier transform from a more engineering perspective, not at least because we often do the transform to throw away unwanted information, like certain frequency components.

In the end it really is just one of many perspectives to look at a function.

[quibono]

> I feel this often gets lost when people approach the Fourier transform from a more engineering perspective, not at least because we often do the transform to throw away unwanted information, like certain frequency components.

That was my problem as well. My first introduction to Fourier transforms was through more of an engineering lens. I remember having trouble with the _inverse_ Fourier transform. I was OK with a Fourier inverse of an already transformed function but I wasn't quite sure what that would mean when applied to a non-transformed, "regular" function.

[sharpneli]

Inverse fourier transform of a non transformed signal gives you basically the fourier transform with some changes (I can't remember which, were the numbers conjugates or something?). Applying it the second time gives you same result as if you'd do the forward direction transform twice.

If you apply fourier transform 4 times you get your original function back. You can think of it as 90 degree rotation. Inverse transform just rotates it in the opposite direction.

The rotation analog is not even too far fetched as fractional fourier transform allows you to do an arbitrary angle rotation.

[araes]

Having never heard of this for the Fourier Transform, needed to read.

F0: original signal

F1: frequency domain signal

F2: reverse time signal

F3: inverse fourier signal

F4: original signal

Also, has further weird applications I've never heard of with "Fractional Fourier Transforms" [1] which can apparently result in smooth smears of time -> frequency domain [2].

[1] https://en.wikipedia.org/wiki/Fractional_Fourier_transform

[2] https://en.wikipedia.org/wiki/File:FracFT_Rec_by_stevencys.j...

[laszlokorte]

See [1] for my visualization of this 4-cycle and the fractional fourier transform

[1]: https://static.laszlokorte.de/frft-cube/

[araes]

Thanks. It's neat being able to visualize them, and the 3D display's actually pretty cool looking for all the different functions.

The Fractional DFT part though, doesn't seem to do anything no matter the function chosen. Firefox 125.

Edit: Nvm, figured it out. Have to visualize from the top down to see the Fractional DFT portion. Haven't seen many visualization systems where each orientation shows a different type of data. Actually a pretty neat idea from a UI perspective.

[takd]

As a related aside, the terms "cepstrum" and the "quefrencies" [1] (c.f. spectrum and frequencies) sound so hilarious that when I first heard about them I was convinced it was some kind of prank.

[1] https://en.wikipedia.org/wiki/Cepstrum

[cmehdy]

This does read like a joke, I had never heard of it either and I'm wondering if many people do use this at all..

Operations on cepstra are labelled quefrency analysis (or quefrency alanysis[1]), liftering, or cepstral analysis. It may be pronounced in the two ways given, the second having the advantage of avoiding confusion with kepstrum.

[aswanson]

It's used in speech signal processing & seismic signal analysis.

[ljosifov]

Almost all speech recognizers until this latest crop (of end-to-end DL NN ASR) operated on cepstral coefficients (and their delta-s and delta-delta-s) as their feature vector.

[pas]

> [...] I wasn't quite sure what that would mean when applied to a non-transformed, "regular" function.

Have you gained some intuition/understanding for this?

I tried a few inputs in WolframAlpha, but unless I manually type in the integral for the inverse transform there's not even a graph :) (and I have no idea whether it's even the same thing without putting a `t` in the exponent and wrapping it in an f(t) = ... )

https://www.wolframalpha.com/input?i=integral+%28sin%28x%29+...

[weinzierl]

Not parent (but GP) and intuition can mean many things but what helped me was keeping in mind:

Every continuous periodic function turns into a discrete aperiodic one when transformed. Works both ways.

Continuous aperiodic stays continuous aperiodic. Discrete periodic stays discrete periodic.

[Salgat]

A fourier transform basically gives you an infinite number of sine waves with different amplitudes/phases at every frequency. If you add them all back together (the inverse fourier transform), you get back your original signal. Audio compression in this case would just be excluding the sine waves that are too high frequency too hear when you add them all back. I always hate how people try to make the fourier transform sound more complex than it actually is (and yes there is more nuance to compression than this, but this is just the basic idea).

[aj7]

The DFT has quite severe limitations that do not appear in the Fourier transform. In particular, the Nyquist criteria that there be zero signal energy for ALL frequencies above half the sampling frequency can only be approximated, and must be accomplished BEFORE sampling, i.e. in the time domain.

[computerfriend]

I wonder how Fourier transformations are taught in engineering courses, because the idea that it could be "lossy" is strange and not obvious to me. It has an inverse after all.

[bisby]

In my engineering courses, Fourier transforms were taught in the context of discrete fourier transforms. Because sampling is a thing that matters (computer audio is discrete data points, not an actual wave).

The Fourier transform of a discrete signal repeats in the frequency domain. For example, [1, -1, 1] could be a sine wave with the exact half of the sampling frequency going from 1 to -1 back to 1 once .... Or it could be a sine wave with double the sampling frequency that is actually going from 1 to -1 to 1 to -1 all within the gap of the first 2 samples. Or it could be 3x the sampling frequency, 4x the sampling frequency, etc. The solution is to only keep the part of the transform that is below the Nyquist limit, because we don't have a sampling rate accurate enough to measure the higher frequencies, so just assuming they dont exist. This also means that if the source signal WAS in fact 4x the sampling frequency, we will see a spike at 1/2 the sampling frequency in the fourier transform, and when we re-create the signal, it will be completely wrong.

So unless you have analog hardware for measuring the Fourier transform (or are working purely in a non-physical mathematical domain, like "i have a sine wave" which can be perfectly represented), you are naturally going to be taking discrete samples of a signal to measure the Fourier transform, which means you are going to be losing any part of the signal that doesn't adhere to sampling rates.

Because my engineering courses were so heavily focused on digital signal processing, when I hear "fourier transform" i immediately think of "discrete fourier transform" and loss is immediately applicable.

[mitthrowaway2]

Don't engineers at least learn to solve the heat equation analytically using Fourier transforms? It wasn't all numerical algorithms, was it?

[woopsn]

It can be lossy and not just because of truncation. Convergence is somewhat finicky in Fourier analysis and was not well understood mathematically before the 60's. Engineers made great use of it anyway.

Remember that it is an integral transform. Basically, any data on a set of vanishing measure can be lost or corrupted. Unfortunately it can even be the case the deviation around some points is unbounded.

[IsTom]

Frequency of a signal and time at which you measure it have an uncertainty principle. https://en.wikipedia.org/wiki/Fourier_transform#Uncertainty_... You either know frequency well, but on a large window or shorten the window and get larger error bars on frequency.

[aswanson]

A square wave would take an infinite number of sinusoidal waves to perfectly reconstruct. To approximate it, one truncates the coefficients. This is done in any engineering application where memory isn't infinite. Which is all of them.

[mkaic]

Of course, a discrete, finite sampling of a square wave at a set of points in time only requires a finite number of coefficients to perfectly reconstruct.

[bisby]

Which is the point. A discrete fourier transform can never recreate a square wave unless you have infinite samples. It can only recreate the finite sampled signal (which is only an approximation of the square wave, and not a real square wave).

This means that sure, the Fourier transform itself isn't lossy (garbage in, garbage out) but Fourier transforms would be used in contexts where loss are introduced. If I have a real perfect square wave, and I want to a take a fourier transform of it, the sampling is going to introduce loss, so to associate sampling losses with the transform itself is fair. Real square wave ran through a DFT program on my computer is going to spit out an approximation of a square wave -- loss.

[davidgay]

The good news of course is that if you were sampling a real signal, then that signal was not actually a perfect square wave. So the fact that you can't (re)construct a perfact square wave is somewhat moot...

[bisby]

Generally yes, but it's a perfectly reasonable assumption that a natural source could generate a signal that is beyond the bounds of what we can record. Any real signal generated by a computer is going to fit within the constraints of what we can generate, but inevitably something like a whale, or a quasar or something will generate a wave that will be lossy.

But also, the question this is all responding to was effectively "why would engineers associate Fourier transforms with loss" and the answer is simply "because the techniques used in calculating most Fourier transforms are going to inherently put a frequency limit and anything beyond that will be lost or show up as an artifact". Engineers work with real world constraints and tend to be hyper aware of those constraints even if they often don't matter.

[kybernetikos]

I think it's the discontinuous jump that causes the problem, and natural signals do sometimes have sudden jumps.